Find the Missing Number in a sorted array

Given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in list. One of the integers is missing in the list. Write an efficient code to find the missing integer. 
Examples:

Input : arr[] = [1, 2, 3, 4, 6, 7, 8]
Output : 5

Input : arr[] = [1, 2, 3, 4, 5, 6, 8, 9]
Output : 7

Naive approach: One Simple solution is to apply methods discussed for finding the missing element in an unsorted array.

 Algorithm

• Create an empty hash table.
• Traverse through the given list of n-1 integers and insert each integer into the hash table.
• Traverse through the range of 1 to n and check whether each integer is present in the hash table or not.
• If any integer is not present in the hash table, then it is the missing integer.

Implementation

C++

#include <iostream>
#include <unordered_set>

using namespace std;

int findMissingNumber(int arr[], int n) {
    unordered_set<int> hashSet;

    // Add all elements of array to hashset
    for (int i = 0; i < n-1; i++) {
        hashSet.insert(arr[i]);
    }

    // Check each integer from 1 to n
    for (int i = 1; i <= n; i++) {
        // If integer is not in hashset, it is the missing integer
        if (hashSet.find(i) == hashSet.end()) {
            return i;
        }
    }

    // If no integer is missing, return n+1
    return n+1;
}

int main() {
    int arr[] = {1, 2, 4, 6, 3, 7, 8};
    int n = sizeof(arr) / sizeof(arr[0]);

    int missingNumber = findMissingNumber(arr, n);
    cout << "Missing number is: " << missingNumber << endl;

    return 0;
}

Java

import java.util.HashSet;

public class Main {
public static int findMissingNumber(int[] arr, int n) {
HashSet<Integer> hashSet = new HashSet<Integer>();
      // Add all elements of array to hashset
    for (int i = 0; i < n-1; i++) {
        hashSet.add(arr[i]);
    }

    // Check each integer from 1 to n
    for (int i = 1; i <= n; i++) {
        // If integer is not in hashset, it is the missing integer
        if (!hashSet.contains(i)) {
            return i;
        }
    }

    // If no integer is missing, return n+1
    return n+1;
}

public static void main(String[] args) {
    int[] arr = {1, 2, 4, 6, 3, 7, 8};
    int n = arr.length;

    int missingNumber = findMissingNumber(arr, n);
    System.out.println("Missing number is: " + missingNumber);
}
}

C#

using System;
using System.Collections.Generic;

class Program
{
    static int FindMissingNumber(int[] arr, int n)
    {
        HashSet<int> hashSet = new HashSet<int>();

        // Add all elements of array to hashset
        for (int i = 0; i < n - 1; i++)
        {
            hashSet.Add(arr[i]);
        }

        // Check each integer from 1 to n
        for (int i = 1; i <= n; i++)
        {
            // If integer is not in hashset, it is the missing integer
            if (!hashSet.Contains(i))
            {
                return i;
            }
        }

        // If no integer is missing, return n+1
        return n + 1;
    }

    static void Main(string[] args)
    {
        int[] arr = { 1, 2, 4, 6, 3, 7, 8 };
        int n = arr.Length;

        int missingNumber = FindMissingNumber(arr, n);
        Console.WriteLine("Missing number is: " + missingNumber);
    }
}

Python

def find_missing_number(arr):
    n = len(arr) + 1
    hash_set = set(arr)

    for i in range(1, n):
        if i not in hash_set:
            return i

    return n

arr = [1, 2, 4, 6, 3, 7, 8]
missing_number = find_missing_number(arr)
print("Missing number is:", missing_number)

Javascript

function findMissingNumber(arr, n) {
  let hashSet = new Set();

  // Add all elements of array to hashset
  for (let i = 0; i < n - 1; i++) {
    hashSet.add(arr[i]);
  }

  // Check each integer from 1 to n
  for (let i = 1; i <= n; i++) {
    // If integer is not in hashset, it is the missing integer
    if (!hashSet.has(i)) {
      return i;
    }
  }

  // If no integer is missing, return n+1
  return n + 1;
}

let arr = [1, 2, 4, 6, 3, 7, 8];
let n = arr.length;

let missingNumber = findMissingNumber(arr, n);
console.log("Missing number is: " + missingNumber);

Output

Missing number is: 5

Time Complexity: O(n), where n is the length of given array
Auxiliary Space: O(n)

Efficient approach: It is based on the divide and conquer algorithm that we have seen in binary search, the concept behind this solution is that the elements appearing before the missing element will have ar[i] – i = 1 and those appearing after the missing element will have ar[i] – i = 2.

Below is the implementation of the above approach:

C++

// A binary search based program to find the
// only missing number in a sorted array of
// distinct elements within limited range.
#include <iostream>
using namespace std;

int search(int ar[], int size)
{
    // Extreme cases
    if (ar[0] != 1)
        return 1;
    if (ar[size - 1] != (size + 1))
        return size + 1;

    int a = 0, b = size - 1;
    int mid;
    while ((b - a) > 1) {
        mid = (a + b) / 2;
        if ((ar[a] - a) != (ar[mid] - mid))
            b = mid;
        else if ((ar[b] - b) != (ar[mid] - mid))
            a = mid;
    }
    return (ar[a] + 1);
}

int main()
{
    int ar[] = { 1, 2, 3, 4, 5, 6, 8 };
    int size = sizeof(ar) / sizeof(ar[0]);
    cout << "Missing number:" << search(ar, size);
}

// This code is contributed by Pushpesh Raj

Java

// A binary search based program
// to find the only missing number
// in a sorted array of distinct
// elements within limited range.
import java.io.*;

class GFG {
    static int search(int ar[], int size)
    {
        // Extreme cases
        if (ar[0] != 1)
            return 1;
        if (ar[size - 1] != (size + 1))
            return size + 1;

        int a = 0, b = size - 1;
        int mid = 0;
        while ((b - a) > 1) {
            mid = (a + b) / 2;
            if ((ar[a] - a) != (ar[mid] - mid))
                b = mid;
            else if ((ar[b] - b) != (ar[mid] - mid))
                a = mid;
        }
        return (ar[a] + 1);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int ar[] = { 1, 2, 3, 4, 5, 6, 8 };
        int size = ar.length;
        System.out.println("Missing number: "
                           + search(ar, size));
    }
}

// This code is contributed
// by inder_verma.

Python3

# A binary search based program to find
# the only missing number in a sorted
# in a sorted array of distinct elements
# within limited range


def search(ar, size):
   # Extreme cases
    if(ar[0] != 1):
        return 1
    if(ar[size-1] != (size+1)):
        return size+1

    a = 0
    b = size - 1
    mid = 0
    while b > a + 1:
        mid = (a + b) // 2
        if (ar[a] - a) != (ar[mid] - mid):
            b = mid
        elif (ar[b] - b) != (ar[mid] - mid):
            a = mid
    return ar[a] + 1


# Driver Code
a = [1, 2, 3, 4, 5, 6, 8]
n = len(a)

print("Missing number:", search(a, n))

# This code is contributed
# by Mohit Kumar

C#

// A binary search based program
// to find the only missing number
// in a sorted array of distinct
// elements within limited range.
using System;

class GFG {
    static int search(int[] ar, int size)
    {
        // Extreme cases
        if (ar[0] != 1)
            return 1;
        if (ar[size - 1] != (size + 1))
            return size + 1;

        int a = 0, b = size - 1;
        int mid = 0;
        while ((b - a) > 1) {
            mid = (a + b) / 2;
            if ((ar[a] - a) != (ar[mid] - mid))
                b = mid;
            else if ((ar[b] - b) != (ar[mid] - mid))
                a = mid;
        }
        return (ar[a] + 1);
    }

    // Driver Code
    static public void Main(String[] args)
    {
        int[] ar = { 1, 2, 3, 4, 5, 6, 8 };
        int size = ar.Length;
        Console.WriteLine("Missing number: "
                          + search(ar, size));
    }
}

// This code is contributed
// by Arnab Kundu

PHP

<?php
// A binary search based program to find the
// only missing number in a sorted array of
// distinct elements within limited range.

function search($ar, $size)
{
   //Extreme cases
    if($ar[0]!=1)
        return 1;
    if($ar[$size-1]!=($size+1))
        return $size+1;

    $a = 0;
    $b = $size - 1;
    $mid;
    while (($b - $a) > 1)
    {
        $mid = (int)(($a + $b) / 2);
        if (($ar[$a] - $a) != ($ar[$mid] -
                                   $mid))
            $b = $mid;
        else if (($ar[$b] - $b) != ($ar[$mid] -
                                        $mid))
            $a = $mid;
    }
    return ($ar[$a] + 1);
}

// Driver Code
$ar = array(1, 2, 3, 4, 5, 6, 8 );
$size = sizeof($ar);
echo "Missing number: ",
     search($ar, $size);

// This code is contributed by ajit.
?>

Javascript

<script>
// A binary search based program
// to find the only missing number
// in a sorted array of distinct
// elements within limited range.

function findMissing(arr) {
  var size = arr.length;
 //Extreme cases
    if(ar[0]!=1)
        return 1;
    if(ar[size-1]!=(size+1))
        return size+1;

  var low = 0;
  var high = arr.length;
  while (low <= high) {
    var mid = Math.floor((low+high)/2);
    if ((arr[mid]-mid === 1) && (arr[mid+1]-(mid+1) === 2)) return arr[mid]+1;
    if (arr[mid]-mid === 1) {
      low = mid+1;
    } else {
      high = mid-1;
    }
  }
  return -1;
}

// Driver Code

let ar = [1, 2, 3, 4, 5, 6, 8];
document.write("Missing number: " +findMissing(ar));

// This code is contributed by mohit kumar 29.
</script>

Output

Missing number:7

Time Complexity: O(log(N)), where N is the length of given array
Auxiliary Space: O(1)